# Theory of Machines Questions and Answers – Worm Gears

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This set of Theory of Machines Multiple Choice Questions & Answers (MCQs) focuses on “Worm Gears”.

1. When large gear reductions are needed _________ gears are used.
a) helical
b) spur
c) worm
d) bevel

Explanation: Worm gears are used where large speed reductions are needed. The horizontal portion of the gear is called as a worm and the assembly of it with the gear is known as a worm gear. Worm can easily turn a gear but a gear cannot turn a worm because of the shallow angle on the worm.

2. The driven gear in the worm gear is a helical gear. True or false?
a) True
b) False

Explanation: The helical gear is driven in the worm gear and the driving element is the screw or the worm. Worm gears are used in transmission of power between two non-parallel and non-intersecting shafts.

3. Which is of these is an advantage of worm gear?
a) It is expensive
b) Has high power losses and low transmission efficiency
c) Produce a lot of heat
d) Used for reducing speed and increasing torque

Explanation: Reduction of speed and increasing the torque is an advantage of worm gear. The rest are the disadvantages of the worm gear. Worm gears are used in gate control mechanisms, hoisting machines, automobile steering mechanisms, lifts, conveyors and presses.
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4. The distance between corresponding points on adjacent teeth measured along the direction of the axis is called ____________
a) joint line
c) axial pitch

Explanation: Axial pitch is the distance between corresponding points on adjacent teeth measured along the direction of the axis. The axial pitch of the worm gear is the same thing as the circular pitch of the helical gear.

5. The distance by which a helix advances along the axis of the gear for one turn around is called _____________
a) joint line
c) axial pitch

Explanation: Lead is the distance by which a helix advances along the axis of the gear for one turn around. The axial pitch is equal to the lead in a single helix and the axial pitch is one half of the lead in a double helix and so on.

6. The angle at which the teeth are inclined to the normal of the axis of rotation is called _______________
a) pitch angle
c) normal angle
d) joint angle

Explanation: Lead angle is the angle at which the teeth are inclined to the normal of the axis of rotation. Lead angle of the worm gear is same as the helix angle of the helical gear. Thus, Ψ2 = λ1.

7. What is the velocity ratio of worm gears?
a) (lπ)/d2
b) (πd2)/l
c) l/(πd2)
d) d2/(lπ)

Explanation: Velocity ratio in worm gears is given as the ratio of the angle turned by the gear to the angle turned by the worm.
Thus, velocity ratio = (2l/d2)/(2π) = l/(πd2).

8. What is the centre distance for the worm gear?
a) (mn/2)(T1 cotλ1 – T2)
b) (mn/2)(T2 cotλ1 + T1)
c) (mn/2)(T2 cotλ1 – T1)
d) (mn/2)(T1 cotλ1 + T2)

Explanation: C = (mn/2)(T1 cotλ1 + T2)
This equation can be derived by using the formula for centre distance of a helical gear which is given as C = (mn/2) ((T1/cos Ψ1) + (T2/cos Ψ2)
As, Ψ2 = λ1, Ψ1 = 90° – λ1.

9. What is the formula to calculate maximum efficiency of a worm gear?
a) (1+sinø)/(1-sinø)
b) (1-sinø)/(1+sinø)
c) (tan(λ1-ø))/tan λ1
d) (tan(λ1+ø))/tan λ1

Explanation: The maximum efficiency of the worm gear is given to be (1-sinø)/(1+sinø), whereas the efficiency of the worm gear is given as (tan(λ1-ø))/tan λ1

10. Find the helix angle of the worm if the pitch of the worm gear is 12 mm and the pitch diameter is 50 mm.
a) 8.687°
b) 11.231°
c) 9.212°
d) 10.319°

Explanation: tan λ1 = Lead / Pitch circumference = 2p/πd1 = 24/50π = 0.1528
λ1 = 8.687°.

11. Find the speed of the gear if the worm is a three start worm rotating at 500 rpm. The gear has 20 teeth.
a) 125 rpm
b) 100 rpm
c) 75 rpm
d) 50 rpm

Explanation: N1/N2 = T2/T1
500/N2 = 20/3
N2 = 75 rpm
Thus, the gear rotates at a speed of 75 rpm.

12. For a two start worm gear having a pitch of 20 mm and a lead angle 12°, find the centre distance if the larger gear has 25 teeth.
a) 148.22 mm
b) 124.93 mm
c) 121.19 mm
d) 109.53 mm

Explanation: C = (mn/2)(T1 cotλ1 + T2)
Therefore, C = (pn/2π)(T1 cotλ1 + T2) = 109.53 mm.

13. Calculate the lead angle of the worm gear for maximum efficiency if θ = 90° and the coefficient of friction is 0.05.
a) 48.21°
b) 42.23°
c) 43.57°
d) 46.43°

Explanation: µ = 0.05°; ø = tan-1(0.05) = 2.862°; θ = 90°
For maximum efficiency, Ψ1 = (θ+ ø)/2 = 92.862/2 = 46.43°
Ψ1 = 90° – λ1 = 46.43°
λ1 = 90° – 46.43° = 43.57°.

14. Find the maximum efficiency if the lead angle is given to be 10° and the coefficient of friction is 0.07.
a) 79.82%
b) 72.23%
c) 76.29%
d) 70.72%

Explanation: λ1 = 10°, ø = tan-1(0.07) = 4°
Efficiency = tan(λ1)/ tan(λ1+ ø) = 0.7072 = 70.72%.

15. Calculate the maximum efficiency of the worm gears which have a friction angle of 0.06.
a) 88.71%
b) 83.23%
c) 89.91%
d) 86.49%

Explanation: ø = tan-1(0.06) = 3.43°
Maximum efficiency = (1-sin ø)/(1+sin ø) = 0.8871 = 88.71%.

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